Compute $\lim_{x\rightarrow 0} \frac{\sin{x}-\arctan{x}}{x(\cos{2x}-1)}$

Question

Using McLaurin expansion for the respective functions I get

$$\sin{x}=x-\frac{x^3}{6}+O(x^5), \\\arctan{x}=x-\frac{x^3}{3}+O(x^5) \\ \cos{t}=1-\frac{t^2}{2}+O(t^4)$$

The expansion for $\cos{t}$ gives

$$x(\cos{2x}-1)= x(1-2x^2+O(16x^4)-1)=-2x^3+xO(16x^4).$$

Combining everything gives the limit

$$\lim_{x\rightarrow 0} \frac{x-\frac{x^3}{6}+O(x^5)-x+\frac{x^3}{3}-O(x^5)}{-2x^3+xO(16x^4)}=\lim_{x\rightarrow 0}\frac{\frac{x^3}{2}+O(x^5)}{x^3(-2+O(16x^4))}=\lim_{x\rightarrow0}\frac{\frac{1}{2}+O(x^5)}{-2+O(16x^4)} \\ =\frac{1}{2}\cdot\frac{1}{-2}=-\frac{1}{4}.$$

Correct answer is $-\frac{1}{12}$.

Questions:

1) Where is my error?

2) Is it correct to multiply an ordo term with $x$ as I've done?

3) How do I know how many terms is enough to expand to when I want to compute a limit?

Write numerator as $$\sin x-x+x-\arctan x$$ and use https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion — lab bhattacharjee, Dec 27 '17 at 14:24

score 4 · Accepted Answer · edited Dec 27 '17 at 14:33

4

The mistake is here: $$-\frac{1}{6}+\frac{1}{3}=\frac{1}{6},$$ but you got $\frac{1}{2}.$

edited Dec 27 '17 at 14:33

José Carlos Santos

427,504

answered Dec 27 '17 at 14:31

Michael Rozenberg

194,933

Ahh basic arithmetic. But Otherwise, are my calculations correct or have I done something mathematically incorrect but accidently got right answer? – Parseval Dec 27 '17 at 14:43
I think, without this mistake your calculations are right. – Michael Rozenberg Dec 27 '17 at 14:44
Thanks a lot mate! – Parseval Dec 27 '17 at 14:45
You are welcome! – Michael Rozenberg Dec 27 '17 at 14:45

score 0 · Answer 2 · answered Dec 27 '17 at 14:29

0

using one times the rules of L'Hospital we get $${\frac {\cos \left( x \right) {x}^{2}+\cos \left( x \right) -1}{ \left( {x}^{2}+1 \right) \left( \cos \left( 2\,x \right) -1-2\,x\sin \left( 2\,x \right) \right) }} $$

answered Dec 27 '17 at 14:29

Dr. Sonnhard Graubner

95,283

score 0 · Answer 3 · answered Dec 27 '17 at 15:18

0

For your 2nd question, your calculation with the $O$ term in the denominator is incorrect, but it does not affect the result. Also, you can remove constant factors (by definition). Here's a correct calculation: $$x(\cos 2x-1)=x\bigl(1-2x^2+O(x^4)-1\bigr)=-2x^3+xO(x^4)=-2x^3+O(x^5). $$

As to your 3rd question, there's no general rule: you only have to obtain ax expansion which is not reduced to its $O$ term. In ither words, you have to obtain an equivalent of the numerator/denominator.

answered Dec 27 '17 at 15:18

Bernard

175,478

But If I have $O(t^4)$ and $t=2x$, won't I get $O((2x)^4)=O(16t^4)?$. One more question, say I have $(x+1)O(x^4)$, how does this simplify? – Parseval Dec 27 '17 at 15:31
$O(16x^4)=O(x^4)$ anyway, so why keep the coefficient? Inside $O$ it is meaningless. As to $(x+1)O(x^4)$, just expand it: $xO(x^4)+O(x^4)=O(x^5)+O(x^4)=O(x^4)$ (near $0$, of course, not near $\infty$). – Bernard Dec 27 '17 at 15:43
Thanks a lot Bernard! Cleard some things up here. – Parseval Dec 27 '17 at 15:44

Compute $\lim_{x\rightarrow 0} \frac{\sin{x}-\arctan{x}}{x(\cos{2x}-1)}$

3 Answers3